Let $z_{1}, z_{2}, ..., z_{n}$ be nonzero complex numbers, with $z_{k}=p_{k}\exp(i\theta_{k})$, where $p_{k}$ is a positive real number and $\theta_{k}$ real.
Can you help me prove that
$\left | \sum_{k=1}^{n}z_{k} \right |^2=\sum_{k=1}^{n}(p_{k})^2+2\sum_{k<l}p_{k}p_{l}\cos(\theta_{k}-\theta_{l})$

1 Answer
1

just multiply
$$(\sum z_j)(\sum \bar{z}_k)=\sum|z_l|^2+\sum_{j\neq k}z_j\bar{z}_k$$
in terms of your $p$ and $\theta$ we have
$$
\sum p_l^2+\sum_{j\neq k} p_jp_ke^{i(\theta_j-\theta_k)}
$$
the cosine terms in $e^{i(\theta_j-\theta_k)}$ double up since cosine is even and the sine terms cancel since sine is odd, i.e.
$$
\cos(\theta_j-\theta_k)+\cos(\theta_k-\theta_j)=2\cos(\theta_j-\theta_k)
$$
and
$$
\sin(\theta_j-\theta_k)+\sin(\theta_k-\theta_j)=0
$$
hence you get the result you quoted

if you want $|\sum z_l|^2=(\sum |z_l|)^2$ then all of the $z_l$ must point in the same direction, in other words the $\theta_l$ must all be the same. plugging that into what we have above, we get
$$
\sum p_l^2+\sum_{j\neq k} p_jp_ke^{i(\theta_j-\theta_k)}=\sum p_l^2+\sum_{j\neq k}p_jp_k=(\sum p_l)^2
$$